1 #include "000库函数.h"
2
3
4 //笨方法，循环，遍历
5 //先将每个字母进行遍历，并给排序，这样就可以确定出有几种字符串，然后就可以开辟空间了
6 //然后给每个字符串进行找位子
7 //笨方法，时间久 864ms ，内存大17.4M
8 class Solution {
9 public:
10 vector<vector<string>> groupAnagrams(vector<string>& strs) {
11 vector<vector<string>>Res;
12 vector<string>st;
13 set<string>s;
14 for (auto a : strs) {
15 sort(a.begin(), a.end());//给每个字符排序
16 s.insert(a);//去除相同字母的组合
17 }
18 st.assign(s.begin(), s.end());//将不相同的字符串储存下来
19 Res.resize(s.size());//给Res开辟空间
20
21 for (auto a : strs) {
22 auto b = a;
23 sort(b.begin(), b.end());//给每个字符排序
24 for (int i = 0; i < st.size(); ++i)
25 if (st[i] == b)
26 Res[i].push_back(a);
27 }
28 return Res;
29
30 }
31 };
32
33 //思路和上面一样，代码简便点，减少了循环次数
34 class Solution {
35 public:
36 vector<vector<string>> groupAnagrams(vector<string>& strs) {
37 vector<vector<string>> res;
38 unordered_map<string, int> mp;
39 for (int i = 0, j = 0; i < strs.size(); i++) {
40 string st = strs[i];
41 sort(st.begin(), st.end());
42 if (mp.find(st) != mp.end()) {
43 res[mp[st]].push_back(strs[i]);
44 }
45 else {
46 vector<string> tmp{ strs[i] };
47 res.push_back(tmp);
48 mp[st] = j++;
49 }
50 }
51 return res;
52 }
53 };
54
55 //用字典最好
56 class Solution {
57 public:
58 vector<vector<string>> groupAnagrams(vector<string>& strs) {
59 vector<vector<string>> res;
60 unordered_map<string, vector<string>> m;
61 for (string str : strs) {
62 string t = str;
63 sort(t.begin(), t.end());
64 m[t].push_back(str);
65 }
66 for (auto a : m) {
67 res.push_back(a.second);
68 }
69 return res;
70 }
71 };
72
73
74 //下面这种解法没有用到排序，我们用一个大小为26的int数组来统计每个单词中字符出现的次数，
75 //然后将int数组转为一个唯一的字符串，跟字符串数组进行映射，这样我们就不用给字符串排序了，
76 //代码如下：
77 class Solution {
78 public:
79 vector<vector<string>> groupAnagrams(vector<string>& strs) {
80 vector<vector<string>> res;
81 unordered_map<string, vector<string>> m;
82 for (string str : strs) {
83 vector<int> cnt(26, 0);
84 string t = "";
85 for (char c : str) ++cnt[c - 'a'];
86 for (int d : cnt) t += to_string(d) + "/";//将数字转换为字母，与相面思想类似，t即为排了序的字符串
87 m[t].push_back(str);
88 }
89 for (auto a : m) {
90 res.push_back(a.second);
91 }
92 return res;
93 }
94 };
95
96 void T049() {
97 Solution s;
98 vector<vector<string>>Res;
99 vector<string>v;
100 v = { "eat", "tea", "tan", "ate", "nat", "bat" };
101 Res = s.groupAnagrams(v);
102 for (auto &a : Res) {
103 for (auto b : a)
104 cout << b << " ";
105 cout << endl;
106 }
107 }
